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Prove that:${c^2} = {(a - b)^2}{\cos ^2}\dfrac{1}{2}C + {(a + b)^2}{\sin ^2}\dfrac{1}{2}C.$

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Hint: Expand the given equation and try to eliminate the trigonometric terms.

Taking R.H.S.,
$ \Rightarrow $${(a - b)^2}{\cos ^2}\dfrac{1}{2}C + {(a + b)^2}{\sin ^2}\dfrac{1}{2}C.$
$ \Rightarrow $$({a^2} + {b^2} - 2ab){\cos ^2}\dfrac{1}{2}C + ({a^2} + {b^2} + 2ab){\sin ^2}\dfrac{1}{2}C$
$ \Rightarrow $${a^2}{\cos ^2}\dfrac{1}{2}C + {b^2}{\cos ^2}\dfrac{1}{2}C - 2ab{\cos ^2}\dfrac{1}{2}C + {a^2}{\sin ^2}\dfrac{1}{2}C + {b^2}{\sin ^2}\dfrac{1}{2}C + 2ab{\sin ^2}\dfrac{1}{2}C$
$ \Rightarrow $\[{a^2}\left( {{{\sin }^2}\dfrac{1}{2}C + {{\cos }^2}\dfrac{1}{2}C} \right) + {b^2}\left( {{{\sin }^2}\dfrac{1}{2}C + {{\cos }^2}\dfrac{1}{2}C} \right) - 2ab\left( {{{\cos }^2}\dfrac{1}{2}C - {{\sin }^2}\dfrac{1}{2}C} \right)\]
We know that, \[\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1\]
Therefore, using this identity, we get,
\[ = {a^2} + {b^2} - 2ab\left( {{{\cos }^2}\dfrac{1}{2}C - {{\sin }^2}\dfrac{1}{2}C} \right)\]
We know the identity,
\[\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = \cos 2\theta \]
Therefore, using this identity, we get,
\[ = {a^2} + {b^2} - 2ab\cos C\]
Now, we know the identity,
$\cos C = \dfrac{{{b^2} + {a^2} - {c^2}}}{{2ab}}$
Therefore, using this identity, we get,
\[ = {a^2} + {b^2} - 2ab\left( {\dfrac{{{b^2} + {a^2} - {c^2}}}{{2ab}}} \right)\]
\[ = {a^2} + {b^2} - ({b^2} + {a^2} - {c^2})\]
\[ = {a^2} + {b^2} - {b^2} - {a^2} + {c^2}\]
\[ = {c^2}\]
That is, LHS=RHS.
So, this is the required solution.

Note: To solve such questions, we should have a good knowledge of various trigonometric identities. We have to analyse each and every step, and have to identify the identity being used to obtain the required solution.

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